Computing $\text{Ext}(\mathbb Z_p,\mathbb Z)$

It is well known that $\text{Ext}(\mathbb Z_p,\mathbb Z)$ is the trivial group, because $\mathbb Z_p$ is projective; this seems to be in contradiction with the Exercise 1.1 in Hilton - Stammbach, pag. 88: the two sequences $$0 \to \mathbb Z \xrightarrow{\mu} \mathbb Z \xrightarrow{\epsilon}\mathbb Z_3 \to 0$$ and $$0 \to \mathbb Z \xrightarrow{\mu} \mathbb Z \xrightarrow{\epsilon'}\mathbb Z_3 \to 0$$ where $\mu$ is the multiplication by 3 in $\mathbb Z$ and $\epsilon(n)=n\pmod{3}$, $\epsilon'(n)=n+1\pmod{3}$ are not equivalent, and then represent different elements in the set of extensions of $\mathbb Z_3$ and $\mathbb Z$... Where am I wrong?

-
Obviously $\mathbb Z_p = \mathbb Z/p\mathbb Z$, where $p$ is a prime number. –  tetrapharmakon Nov 22 '10 at 21:09
This is not obvious; Z_p (in my experience) more frequently refers to the p-adic integers. –  Qiaochu Yuan Nov 22 '10 at 21:11
Yes, "obviously" was a lapsus linguae. p-adic integers are $\mathbb Z_{(p)}$. –  tetrapharmakon Nov 22 '10 at 21:15
Right, but the additive group of the $p$-adic integers isn't a projective $\mathbb{Z}$-module either. Projective modules over a PID are free. But a nonzero free abelian group stays nonzero upon tensoring with $\mathbb{Z}/\ell\mathbb{Z}$ for all primes $\ell$, unlike $\mathbb{Z}_p$. It is flat though... –  Pete L. Clark Nov 22 '10 at 21:35
No, "obviously" not necessarily. $\mathbb{Z}_{(p)}$ often refers to the localization of $\mathbb{Z}$ at the prime $(p)$, that is, all rationals which can be expressed with denominator prime to $p$. There are too many conflicting notations in this particular area to assume that anything is "obvious"! –  Arturo Magidin Nov 22 '10 at 21:54

Well, $\mathbb Z_p=\mathbb Z/p\mathbb Z$ certainly isn't projective as a $\mathbb Z$-module.

Added Now $\mathrm{Ext}(\mathbb Z/p\mathbb Z,\mathbb Z)$ is cyclic of order $p$. The zero element corresponds to the split extension of $\mathbb Z$ by $\mathbb Z_p$. The other elements correspond to extensions $$0\to \mathbb Z\to \mathbb Z\to \mathbb Z/p\mathbb Z\to0.\qquad\qquad\qquad(*)$$ In this exact sequence the surjection can be taken to map $1\in \mathbb Z$ to any given nonzero $a\in\mathbb Z/p\mathbb Z$. Thus there are $p-1$ non-isomorphic extensions looking like $(*)$.

-
No, it isn't, you're right (and I'm completely wrong). Then, how can I do to compute it starting from the definition of classes of extensions of the given sequence? –  tetrapharmakon Nov 22 '10 at 21:25

Here is one way to see why $\Bbb{Z}/p\Bbb{Z}$ isn't projective. Suppose you take $f$ to be the projection $f : \Bbb{Z} \to \Bbb{Z}/p\Bbb{Z}$. Now if $\Bbb{Z}/p\Bbb{Z}$ is projective then for every map $g : \Bbb{Z}/p\Bbb{Z} \to \Bbb{Z}$ we will get a map $h : \Bbb{Z}/p\Bbb{Z} \to \Bbb{Z}$ such that

$$f \circ h = g.$$

However the problem now is that there are no homomorphisms from $\Bbb{Z}/p\Bbb{Z}$ to $\Bbb{Z}$ except the trivial homomorphism!

IMO to compute the first Ext group in your problem, the easiest would be to look at the ses

$$0 \to \Bbb{Z} \stackrel{p\cdot}{\to} \Bbb{Z} \to \Bbb{Z}/p\Bbb{Z} \to 0$$

that gives rise to the following LES in Ext:

$$0\to \textrm{Hom}(\Bbb{Z}/p\Bbb{Z},\Bbb{Z}) \to \textrm{Hom}(\Bbb{Z},\Bbb{Z}) \stackrel{p\cdot}{\to} \textrm{Hom}(\Bbb{Z},\Bbb{Z}) \to \textrm{Ext}^1(\Bbb{Z}/p\Bbb{Z},\Bbb{Z}) \to 0 \to 0\ldots$$

We have the zeros at the end because $\Bbb{Z}$ is a free abelian group and hence is projective. Since $\textrm{Hom}(\Bbb{Z},\Bbb{Z}) \cong \Bbb{Z}$ by the first isomorphism theorem we get

$$\textrm{Ext}^1(\Bbb{Z}/p\Bbb{Z},\Bbb{Z}) \cong \Bbb{Z}/p\Bbb{Z}.$$

-